Open Access

The upper bound estimate of the number of integer points on elliptic curves y 2 = x 3 + p 2 r x

Journal of Inequalities and Applications20142014:104

https://doi.org/10.1186/1029-242X-2014-104

Received: 22 January 2014

Accepted: 20 February 2014

Published: 4 March 2014

Abstract

Let p be a fixed prime and r be a fixed positive integer. Further let N ( p 2 r ) denote the number of pairs of integer points ( x , ± y ) on the elliptic curve E : y 2 = x 3 + p 2 r x with y > 0 . Using some properties of Diophantine equations, we give a sharper upper bound estimate for N ( p 2 r ) . That is, we prove that N ( p 2 r ) 1 , except with N ( 17 2 ( 2 s + 1 ) ) = 2 , where s is a nonnegative integer.

MSC:11G05, 11Y50.

Keywords

elliptic curveinteger pointDiophantine equation

1 Introduction

Let , be the sets of all integers and positive integers, respectively. Let p be a fixed prime and k be a fixed positive integer. Recently, the integer points ( x , y ) on the elliptic curve
y 2 = x 3 + p k x
(1.1)
have been investigated in many papers (see [1, 2] and [3]). In this paper we deal with the number of integer points on (1.1) for even k. Then (1.1) can be rewritten as
y 2 = x 3 + p 2 r x ,
(1.2)

where r is a positive integer.

An integer point ( x , y ) on (1.2) is called trivial or non-trivial according to whether y = 0 or not. Obviously, (1.2) has only the trivial integer point ( x , y ) = ( 0 , 0 ) . Notice that if ( x , y ) is a non-trivial integer point on (1.2), then ( x , y ) is also. Therefore, ( x , y ) along with ( x , y ) are called by a pair of non-trivial integer points and denoted by ( x , ± y ) , where y > 0 . For any positive integer n, let
u ( n ) = 1 2 ( α n + β n ) , v ( n ) = 1 2 2 ( α n β n ) ,
(1.3)
where
α = 3 + 2 2 , β = 3 2 2 .
(1.4)

Using some properties of Diophantine equations, we give a sharper upper bound estimate for N ( p 2 r ) , the number of pairs of non-trivial integer points ( x , ± y ) on (1.2). That is, we shall prove the following results.

Theorem 1.1 All non-trivial integer points on (1.2) are given as follows.
  1. (i)

    p = u ( 2 m ) , r = 2 s + 1 , ( x , ± y ) = ( p 2 s v 2 ( 2 m ) , ± p 3 s v ( 2 m ) ( v 2 ( 2 m ) + 1 ) ) , where m, s are nonnegative integers.

     
  2. (ii)
    p 1 ( mod 8 ) , r = 2 s + 1 , ( x , ± y ) = ( p 2 s + 1 X 2 , ± p 3 s + 2 X Y ) , where s is a nonnegative integer, ( X , Y ) is a solution of the equation
    X 4 p Y 2 = 1 , X , Y N .
    (1.5)
     

Theorem 1.2 Let p be an odd prime, r be a positive integer. Then for any nonnegative integer s, we have N ( p 2 r ) 1 , except with N ( 17 2 ( 2 s + 1 ) ) = 2 . Moreover, if p 1 ( mod 8 ) , then N ( p 2 r ) = 0 , except with N ( 3 2 ( 2 s + 1 ) ) = 1 .

2 Lemmas

Lemma 2.1 ([[4], Theorem 244])

Every solution ( u , v ) of the equation
u 2 2 v 2 = 1 , u , v N
(2.1)

can be expressed as ( u , v ) = ( u ( n ) , v ( n ) ) , where n is a positive integer.

Lemma 2.2 If p = u ( n ) , then n = 2 m , where m is a nonnegative integer.

Proof Assume that n has an odd divisor d with d > 1 . Then we have either u ( 1 ) | u ( n ) and 1 < u ( 1 ) < u ( n ) or u ( n / d ) | u ( n ) and 1 < u ( n / d ) < u ( n ) . Therefore, since p is a prime, it is impossible. Thus, we get n = 2 m . The lemma is proved. □

Any fixed positive integer a can be uniquely expressed as a = b c 2 , where b, c are positive integers with b is square free. Then b is called the quadratfrei of a and denoted by Q ( a ) .

Lemma 2.3 For any positive integer m, we have 3 | Q ( v ( 2 m ) ) .

Proof By (1.3) and (1.4), we get
v ( 2 m ) = 2 m + 1 i = 0 m 1 u ( 2 i )
(2.2)
and
u ( 2 i ) = 2 u 2 ( 2 i 1 ) 1 , i N .
(2.3)

Since ( 2 / 3 ) = 1 , where ( 2 / 3 ) is the Legendre symbol, we see from (2.3) that 3 u ( 2 i ) for i 1 . Therefore, since u ( 1 ) = 3 , by (2.2), we obtain 3 v ( 2 m ) . It implies that 3 | Q ( v ( 2 m ) ) . The lemma is proved. □

Let D be a non-square positive integer. It is a well known fact that if the equation
U 2 D V 2 = 1 , U , V N
(2.4)
has solutions ( U , V ) , then it has a unique solution ( U 1 , V 1 ) such that U 1 + V 1 D U + V D , where ( U , V ) through all solutions of (2.4). For any odd positive integer l, let
U ( l ) + V ( l ) D = ( U 1 + V 1 D ) l .

Then ( U , V ) = ( U ( l ) , V ( l ) ) ( l = 1 , 3 , ) are all solutions of (2.4).

Lemma 2.4 ([[5], Theorem 1])

The equation
X 4 D Y 2 = 1 , X , Y N
(2.5)

has at most one solution ( X , Y ) . Moreover, if the solution ( X , Y ) exists, then ( X 2 , Y ) = ( U ( l ) , V ( l ) ) , where l = Q ( U 1 ) .

Lemma 2.5 ([[5], Theorem 3])

If 3 | Q ( U 1 ) , then (2.5) has no solutions ( X , Y ) .

Lemma 2.6 If p = u ( 2 m ) , where m is a positive integer with m > 1 , then (1.5) has no solutions ( X , Y ) .

Proof Since p = u ( 2 m ) with m > 1 , by (2.3), we have
p = 2 u 2 ( 2 m 1 ) 1 = 4 v 2 ( 2 m 1 ) + 1 .
(2.6)
We see from (2.6) that the equation
U 2 p V 2 = 1 , U , V N
(2.7)

has solution ( U , V ) and its fundamental solution is ( U 1 , V 1 ) = ( 2 v ( 2 m 1 ) , 1 ) . Further, since m 1 1 , by Lemma 2.3, we have 3 | Q ( v ( 2 m 1 ) ) . Hence, we get 3 | Q ( U 1 ) = Q ( 2 v ( 2 m 1 ) ) . Therefore, by Lemma 2.5, the lemma is proved. □

Lemma 2.7 ([6])

The equation
2 X 2 + 1 = Y n , X , Y , n N , n > 3
(2.8)

has no solutions ( X , Y , n ) .

Lemma 2.8 The equation
X 2 Y 4 = p 2 n , X , Y , n N , gcd ( X , Y ) = 1
(2.9)

has only the solutions ( p , X , Y , n ) = ( u ( 2 m ) , v 2 ( 2 m ) + 1 , v ( 2 m ) , 1 ) , where m is a nonnegative integer.

Proof Assume that ( p , X , Y , n ) is a solution of (2.9). If p = 2 , since gcd ( X , Y ) = 1 , then we have 2 X Y , gcd ( X + Y 2 , X Y 2 ) = 2 , X + Y 2 = 2 2 n 1 , X Y 2 = 2 and Y 2 = 2 2 n 2 1 . But since Y 2 + 1 is not a square, it is impossible.

If p is an odd prime, then we have gcd ( X + Y 2 , X Y 2 ) = 1 , and by (2.9), we get X + Y 2 = p 2 n , X Y 2 = 1 ,
2 X = p 2 n + 1
(2.10)
and
2 Y 2 = p 2 n 1 .
(2.11)
By Lemma 2.7, we get from (2.11) that n = 1 and
p 2 2 Y 2 = 1 .
(2.12)
Further, applying Lemma 2.1 to (2.12) yields
p = u ( n ) , Y = v ( n ) , n N .
(2.13)

Further, by Lemma 2.2, we see from the first equality of (2.13) that n = 2 m . Thus, by (2.10) and (2.13), the lemma is proved. □

3 Proof of Theorem 1.1

Assume that ( x , ± y ) is a pair of non-trivial integer points on (1.2). Since y > 0 , we have x > 0 and x can be expressed as
x = p t z , t Z , t 0 , z N , p z .
(3.1)
Substituting (3.1) into (1.2) yields
p t z ( p 2 t z 2 + p 2 r ) = y 2 .
(3.2)
We first consider the case that r > t . By (3.2), we have
p 3 t z ( z 2 + p 2 r 2 t ) = y 2 .
(3.3)
Since p z , we have p z 2 + p 2 r 2 t and gcd ( z , z 2 + p 2 k 2 r ) = 1 . Hence by (3.3), we get
t = 2 s , z = f 2 , z 2 + p 2 r 2 t = g 2 , y = p 3 s f g , s Z , s 0 , f , g N , gcd ( f , g ) = 1 ,
(3.4)
whence we obtain
g 2 f 4 = p 2 r 4 s .
(3.5)
Applying Lemma 2.8 to (3.5) yields
p = u ( 2 m ) , 2 r 4 s = 2 , f = v ( 2 m ) , g = v 2 ( 2 m ) + 1 , m Z , m 0 .
(3.6)

Therefore, by (3.1), (3.4), and (3.6), the integer points of type (i) are given.

We next consider the case that r = t . Then we have
p 3 r z ( z 2 + 1 ) = y 2 .
(3.7)
Since p z , gcd ( z , z 2 + 1 ) = 1 and z 2 + 1 is not a square, we see from (3.7) that
r = 2 s + 1 , z = f 2 , z 2 + 1 = p g 2 , y = p 3 s + 2 f g , s Z , s 0 , f , g N , gcd ( f , g ) = 1 .
(3.8)
By (3.8), we get
f 4 p g 2 = 1 .
(3.9)

It implies that ( X , Y ) = ( f , g ) is a solution of (1.5). Therefore, by (3.1) and (3.8), we obtain the integer points of type (ii).

We finally consider the case that r < t . Then we have
p t + 2 r z ( p 2 t 2 r z 2 + 1 ) = y 2 .
(3.10)

Since p z ( p 2 t 2 r z 2 + 1 ) and gcd ( z , p 2 t 2 r z 2 + 1 ) = 1 , we see from (3.10) that p 2 t 2 r z 2 + 1 is a square, a contradiction.

To sum up, the theorem is proved.

4 Proof of Theorem 1.2

By (2.3), if p = u ( 2 m ) with m 1 , then p 1 ( mod 8 ) . Therefore, by Theorem 1.1, if p 1 ( mod 8 ) , then (1.2) has only the non-trivial integer point
p = 3 , r = 2 s + 1 , ( x , ± y ) = ( 3 2 s 4 , ± 3 3 s 10 ) .
(4.1)

It implies that the theorem is true for p 1 ( mod 8 ) .

For p 1 ( mod 8 ) , let N 1 and N 2 denote the number of pairs of non-trivial integer points of types (i) and (ii) in Theorem 1.1, respectively. Obviously, we have
N ( p 2 r ) = N 1 + N 2
(4.2)
and N 1 1 . By Lemma 2.4, we get N 2 1 . Hence, by (4.2), we have N ( p 2 r ) 2 for p 1 ( mod 8 ) . Since u ( 2 ) = 17 and (1.5) has the solution ( X , Y ) = ( 2 , 1 ) for p = 17 , by Theorem 1.1, we get
p = 17 , r = 2 s + 1 , ( x , ± y ) = ( 17 2 s 144 , ± 17 3 s 1 , 740 ) and ( 17 2 s + 1 4 , ± 17 3 s + 2 2 )
(4.3)

and N ( 17 2 ( 2 s + 1 ) ) = 2 . However, by Lemma 2.6, if p = u ( 2 m ) with m > 1 , then N 2 = 0 . Therefore, by (4.2), if p 1 ( mod 8 ) , then N ( p 2 r ) 1 , except with N ( 17 2 ( 2 s + 1 ) ) = 2 . The theorem is proved.

Declarations

Acknowledgements

The authors would like to thank the referees for their very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P.E.D. (2013JK0573) and N.S.F. (11371291) of P.R. China.

Authors’ Affiliations

(1)
School of Mathematics and Computer Engineering, University of Arts and Science
(2)
Department of Mathematics, Northwest University

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Copyright

© Zhang and Li; licensee Springer. 2014

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